高一物理(有关小船过河)一条小船保持对水恒定的速度过河.若船头垂直河岸划行,经10min到达正对岸下游120m处;若船头指向与上游河岸成Θ角划行,经12.5min到达正对岸,则水速u=?Θ=?船对
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/20 07:00:36
![高一物理(有关小船过河)一条小船保持对水恒定的速度过河.若船头垂直河岸划行,经10min到达正对岸下游120m处;若船头指向与上游河岸成Θ角划行,经12.5min到达正对岸,则水速u=?Θ=?船对](/uploads/image/z/9502762-58-2.jpg?t=%E9%AB%98%E4%B8%80%E7%89%A9%E7%90%86%EF%BC%88%E6%9C%89%E5%85%B3%E5%B0%8F%E8%88%B9%E8%BF%87%E6%B2%B3%EF%BC%89%E4%B8%80%E6%9D%A1%E5%B0%8F%E8%88%B9%E4%BF%9D%E6%8C%81%E5%AF%B9%E6%B0%B4%E6%81%92%E5%AE%9A%E7%9A%84%E9%80%9F%E5%BA%A6%E8%BF%87%E6%B2%B3%EF%BC%8E%E8%8B%A5%E8%88%B9%E5%A4%B4%E5%9E%82%E7%9B%B4%E6%B2%B3%E5%B2%B8%E5%88%92%E8%A1%8C%2C%E7%BB%8F10min%E5%88%B0%E8%BE%BE%E6%AD%A3%E5%AF%B9%E5%B2%B8%E4%B8%8B%E6%B8%B8120m%E5%A4%84%EF%BC%9B%E8%8B%A5%E8%88%B9%E5%A4%B4%E6%8C%87%E5%90%91%E4%B8%8E%E4%B8%8A%E6%B8%B8%E6%B2%B3%E5%B2%B8%E6%88%90%CE%98%E8%A7%92%E5%88%92%E8%A1%8C%2C%E7%BB%8F12.5min%E5%88%B0%E8%BE%BE%E6%AD%A3%E5%AF%B9%E5%B2%B8%2C%E5%88%99%E6%B0%B4%E9%80%9Fu%EF%BC%9D%3F%CE%98%EF%BC%9D%3F%E8%88%B9%E5%AF%B9)
高一物理(有关小船过河)一条小船保持对水恒定的速度过河.若船头垂直河岸划行,经10min到达正对岸下游120m处;若船头指向与上游河岸成Θ角划行,经12.5min到达正对岸,则水速u=?Θ=?船对
高一物理(有关小船过河)
一条小船保持对水恒定的速度过河.若船头垂直河岸划行,经10min到达正对岸下游120m处;若船头指向与上游河岸成Θ角划行,经12.5min到达正对岸,则水速u=?Θ=?船对水的速度v=?河宽l=?
高一物理(有关小船过河)一条小船保持对水恒定的速度过河.若船头垂直河岸划行,经10min到达正对岸下游120m处;若船头指向与上游河岸成Θ角划行,经12.5min到达正对岸,则水速u=?Θ=?船对
设船速度V,垂直河岸时V=I/t =I/10min ①
u=120m/t=120m/10min=0.2m/s即水流速度u=0.2m/s
与河岸成角度a时(你那符号我打不出来)
船垂直与河岸的速度V2=I/t2=I/12.5min ②
联立①②
V/V2=t2/t=12.5min/10min=5:4
故sin a=V2/V=4:5
所以cos a=3:5
因为成a角时,平行河岸的速度V3=v*sin a,经12.5min到达正对岸,所以V3=u=0.2m/s
所以V=V3/cos a=0.2/(3/5)m/s=0.333m/s
所以河宽I=Vt=0.333m/s *10min=199.8m
画力学三角就可以了,很简单
1.u*t1=s=120m;u=s/t1=120m/600s=0.2m/s
2.L=V*t1=(根号V^2-u^2)*t2解得V=0.33m/s
3.L=V*t1=0.33*600=198m
4.cosΘ=u/V=0.2/0.33=20/33